Derive the expression for a body projected horizontally.

Explanation

(i) At any instant t , the projectile has velocity components along both x - axis and y - axis .

(ii) The resultant of these two components gives the velocity of the projectile at that instant t , as shown in Figure
(iii) The velocity component at any t along horizontal (x - axis) is `v_(x)=u_(x)+a_(x)t`
since, `u_(x)=u, a_(x)=0,` we get
`v_(x)=u.`
(iv) The component of velocity along vertical direction (y - axis) is `v_(y)=u_(y)+a_(y)t`
Since, `u_(y)=0, a_(y)=g,` we get `v_(y)=g t`
(v) Hence the velocity of the particle at any instant is `vecv=uhati+g t hatj`
(vi) The speed of the particle at any instant t is given by
`therefore" "v=sqrt(v_(x)^(2)+v_(y)^(2)), v=sqrt(u^(2)+g^(2)t^(2))`